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A Real Value of X Satisfies the Equation 3 − 4 I X 3 + 4 I X = a − I B ( a , B ∈ R ) , I F a 2 + B 2 = - Mathematics

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Question

A real value of x satisfies the equation  \[\frac{3 - 4ix}{3 + 4ix} = a - ib (a, b \in \mathbb{R}), if a^2 + b^2 =\]

Options

  • 1

  • -1

  • 2

  • -2

MCQ
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Solution

\[a - ib = \frac{3 - 4ix}{3 + 4ix}\]

\[ = \frac{3 - 4ix}{3 + 4ix} \times \frac{3 - 4ix}{3 - 4ix}\]

\[ = \frac{9 + 16 x^2 i^2 - 24xi}{9 - 16 x^2 i^2}\]

\[ = \frac{\left( 9 - 16 x^2 \right) - i\left( 24x \right)}{9 + 16 x^2}\]

\[ \Rightarrow \left| a - ib \right|^2 = \left| \frac{\left( 9 - 16 x^2 \right) - i\left( 24x \right)}{9 + 16 x^2} \right|^2 \]

\[ \Rightarrow a^2 + b^2 = \frac{\left( 9 - 16 x^2 \right)^2 + \left( 24x \right)^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = \frac{81 + 256 x^4 - 288 x^2 + 576 x^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = \frac{81 + 256 x^4 + 288 x^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = \frac{\left( 9 + 16 x^2 \right)^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = 1\]

Hence, the correct option is (a).

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Chapter 13: Complex Numbers - Exercise 13.6 [Page 66]

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RD Sharma Mathematics [English] Class 11
Chapter 13 Complex Numbers
Exercise 13.6 | Q 39 | Page 66

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